Coordinate Geometry

Class 10 . Mathematics . Mathematics . Coordinate Geometry

Chapter 7: Coordinate Geometry

Ch 7

Mathematics

Class 10

Lecture 1

Introduction & Distance Formula

Reference video · Lecture 1Introduction & Distance Formula

Introduction to Coordinate Geometry

Distance Formula

Distance formula

For $P(x_1,y_1)$ and $Q(x_2,y_2)$, the horizontal and vertical differences are $x_2-x_1$ and $y_2-y_1$. Pythagoras' theorem gives the distance; take the non-negative square root.

$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Distance between two points

$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$

Reversing the order of the points gives the same distance.

Distance from the origin

$$OP=\sqrt{x^2+y^2}$$

Set one point equal to the origin $(0,0)$.

Distance between points on one axis

$$\text{On the }x\text{-axis: }|x_2-x_1|;\qquad \text{on the }y\text{-axis: }|y_2-y_1|$$

Use the absolute difference of the coordinates that vary.

Finding the distance

Write the points as $P(x_1,y_1)$ and $Q(x_2,y_2)$.
Find the differences between corresponding coordinates.
$$x_2-x_1,\quad y_2-y_1$$
Square the differences, add them, and take the positive square root.
$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Worked example

Find the distance between $P(4,6)$ and $Q(6,8)$.

Find the distance between $P(4,6)$ and $Q(6,8)$.
Substitute the coordinate differences into the distance formula.
$$PQ=\sqrt{(6-4)^2+(8-6)^2}$$
Square, add, and simplify.
$$PQ=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$$

Answer: $2\sqrt{2}$ units

⚠ Common mistake

Do not treat a coordinate difference as a distance without considering its sign; square the differences and use the non-negative square root.

Using Distance to Study Points and Figures

Converse of Pythagoras' theorem

If the squares of the two shorter sides add to the square of the longest side, the triangle is right-angled.

$$a^2+b^2=c^2$$

Definition

Equidistant point — A point is equidistant from two fixed points when its distances to them are equal; all such points lie on their perpendicular bisector.

Testing a shape or point relationship

Use the distance formula to calculate the required side lengths.
Compare lengths to test equal sides, diagonals, or a collinearity relation.
For a right angle, use the converse of Pythagoras' theorem; for equidistance, set the squared distances equal and simplify.

Practice

Do $P(3,2)$, $Q(-2,-3)$ and $R(2,3)$ form a triangle? If so, classify it.

Hint: Find the three distances, check the triangle inequality, then compare the squared lengths.

Key step: Check whether $PQ^2+PR^2=QR^2$.

Solution: $PQ=\sqrt{50}$, $PR=\sqrt{2}$, and $QR=\sqrt{52}$. Since $PQ+PR>QR$ and $PQ^2+PR^2=QR^2$, they form a right triangle, right-angled at $P$.

Practice

Show that $A(1,7)$, $B(4,2)$, $C(-1,-1)$ and $D(-4,4)$ are the vertices of a square.

Hint: Find all four sides and the diagonals, or use three sides and a right-angle test.

Key step: Show that all sides have length $\sqrt{34}$ and both diagonals have length $\sqrt{68}$.

Solution: $AB=BC=CD=DA=\sqrt{34}$ and $AC=BD=\sqrt{68}$; hence $ABCD$ is a square.

Practice

(a) Are $A(3,1)$, $B(6,4)$ and $C(8,6)$ collinear? (b) Find the relation for points equidistant from $(7,1)$ and $(3,5)$. (c) Find the point on the $y$-axis equidistant from $(6,5)$ and $(-4,3)$.

Hint: For (a), compare $AB+BC$ with $AC$. For (b) and (c), set squared distances equal; use $(0,y)$ for a point on the $y$-axis.

Key step: Use the distance formula for (a); cancel the squared terms after equating distances in (b) and (c).

Solution: (a) $AB=3\sqrt{2}$, $BC=2\sqrt{2}$, and $AC=5\sqrt{2}$, so the points are collinear. (b) Equating squared distances gives $(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$, hence $x-y=2$; this is the perpendicular bisector. (c) Let $P=(0,y)$. Equating squared distances gives $36+(y-5)^2=16+(y-3)^2$, so $y=9$ and $P=(0,9)$.

★ Remember

Distance depends on the squared differences of corresponding coordinates and is always non-negative.
Use distance comparisons to test collinearity, right triangles, squares, and equidistance.
Points equidistant from two fixed points lie on the perpendicular bisector of the segment between them.

Quick recap

The abscissa is the $x$-coordinate and the ordinate is the $y$-coordinate.
Points on the $x$-axis have form $(x,0)$; points on the $y$-axis have form $(0,y)$.
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
The distance from $(0,0)$ to $(x,y)$ is $\sqrt{x^2+y^2}$.
Distance comparisons can establish collinearity, right angles, squares, and equidistance.
The chapter also studies coordinates of a point dividing a segment in a given ratio.

Lecture 2

Section Formula

Reference video · Lecture 2Section Formula

Internal Division of a Line Segment

Definition

Section formula — The section formula gives the coordinates of a point dividing the line segment between two known points in a specified ratio.

Key idea

Internal division: $P$ divides $AB$ internally when it lies between $A$ and $B$.
Ratio convention: If $PA:PB=m_1:m_2$, pair $m_1$ with the coordinates of $B$ and $m_2$ with those of $A$.
Derivation: Perpendiculars to the axes form similar triangles, so the horizontal and vertical coordinate changes are divided in the same ratio.
External division: A point on line $AB$ but outside segment $AB$ divides externally; its formula is not covered here.

Point P lies between the origin and B(36,15). — Point $P$ lies between the origin and $B(36,15)$.

Relay-tower example

Given: Let $A=O=(0,0)$, $B=(36,15)$, and $OP:PB=1:2$.
Coordinate ratios: If $P=(x,y)$, then $\frac{x}{36-x}=\frac{1}{2}$ and $\frac{y}{15-y}=\frac{1}{2}$.
Result: Solving gives $x=12$ and $y=5$, so the tower is at $P=(12,5)$.

Points A, P, and B lie on one line, with P between the endpoints. — Points $A$, $P$, and $B$ lie on one line, with $P$ between the endpoints.

Section formula for internal division

If $P(x,y)$ divides the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $PA:PB=m_1:m_2$, then its coordinates are weighted averages of the endpoint coordinates.

$$\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$$

Ratio $k:1$ and midpoint

$$P=\left(\frac{kx_2+x_1}{k+1},\frac{ky_2+y_1}{k+1}\right),\qquad M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

For ratio $k:1$, set $m_1=k$ and $m_2=1$. The midpoint is the special case $1:1$.

Finding a point that divides a segment

Identify the endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$, and the ratio $PA:PB=m_1:m_2$.
Substitute the endpoint coordinates and ratio terms into the section formula.
$$\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$$
Simplify both coordinates and check that the point lies between the endpoints for internal division.

Worked example

Find the coordinates of the point dividing the segment joining $(4,-3)$ and $(8,5)$ internally in the ratio $3:1$.

Find the point dividing the segment from $(4,-3)$ to $(8,5)$ internally in the ratio $3:1$.
Use $m_1=3$, $m_2=1$, $A=(4,-3)$, and $B=(8,5)$ in the section formula.
$$P=\left(\frac{3(8)+1(4)}{3+1},\frac{3(5)+1(-3)}{3+1}\right)$$
Simplify each coordinate.
$$P=\left(\frac{28}{4},\frac{12}{4}\right)=(7,3)$$

Answer: The required point is $(7,3)$.

Applications of the Section Formula

Useful applications

Trisection: The two points dividing a segment into three equal parts divide it in ratios $1:2$ and $2:1$ from endpoint $A$.
Point on the $y$-axis: Its $x$-coordinate is $0$; set the section-formula $x$-coordinate equal to $0$.
Parallelogram: Its diagonals bisect each other, so the diagonals have the same midpoint.

The two points of trisection divide AB into three equal parts. — The two points of trisection divide $AB$ into three equal parts.

Practice

In what ratio does $(-4,6)$ divide the segment joining $A(-6,10)$ and $B(3,-8)$ internally?

Hint: Let $PA:PB=m_1:m_2$ and equate the section-formula $x$-coordinate to $-4$.

Key step: $-4=\frac{3m_1-6m_2}{m_1+m_2}$ gives $7m_1=2m_2$.

Solution: Thus $m_1:m_2=2:7$. Check the $y$-coordinate: $\frac{-8(2)+10(7)}{2+7}=6$.

Practice

Find the trisection points of the segment joining $A(2,-2)$ and $B(-7,4)$.

Hint: Use the ratios $1:2$ and $2:1$ for the two points.

Key step: Apply the section formula first with $(m_1,m_2)=(1,2)$ and then with $(m_1,m_2)=(2,1)$.

Solution: For ratio $1:2$, $P=\left(\frac{-7+2(2)}{3},\frac{4+2(-2)}{3}\right)=(-1,0)$. For ratio $2:1$, $Q=\left(\frac{2(-7)+2}{3},\frac{2(4)-2}{3}\right)=(-4,2)$.

Practice

(a) Find where the $y$-axis divides the segment joining $(5,-6)$ and $(-1,-4)$. (b) If $A(6,1)$, $B(8,2)$, $C(9,4)$, and $D(p,3)$ are consecutive vertices of a parallelogram, find $p$.

Hint: (a) Set the section-formula $x$-coordinate to $0$. (b) Equate the midpoints of diagonals $AC$ and $BD$.

Key step: (a) For ratio $k:1$, solve $\frac{-k+5}{k+1}=0$. (b) Equate the diagonal midpoints: $\frac{15}{2}=\frac{8+p}{2}$.

Solution: (a) $k=5$, so the ratio is $5:1$ and the intersection is $\left(0,-\frac{13}{3}\right)$. (b) $p=7$.

Distance formulas (chapter recap)

$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\qquad OP=\sqrt{x^2+y^2}$$

For points $P(x_1,y_1)$ and $Q(x_2,y_2)$, the distance formula gives $PQ$. Setting one point to the origin gives the distance from the origin.

⚠ Common mistake

Do not pair $m_1$ with the coordinates of $A$: when $PA:PB=m_1:m_2$, the numerator uses $m_1(x_2,y_2)+m_2(x_1,y_1)$.

Quick recap

For internal division with $PA:PB=m_1:m_2$, use $P=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$.
For ratio $k:1$, set $m_1=k$ and $m_2=1$; the midpoint is the special case $1:1$.
The midpoint coordinates are the averages of the corresponding endpoint coordinates.
Trisection points divide a segment in ratios $1:2$ and $2:1$ from the first endpoint.
A point on the $y$-axis has $x=0$; a parallelogram's diagonals have the same midpoint.
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$; the distance from the origin to $(x,y)$ is $\sqrt{x^2+y^2}$.

Formula sheet

Distance between two points

$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$

Distance from the origin

$$OP=\sqrt{x^2+y^2}$$

Distance between points on one axis

$$\text{On the }x\text{-axis: }|x_2-x_1|;\qquad \text{on the }y\text{-axis: }|y_2-y_1|$$

Ratio $k:1$ and midpoint

$$P=\left(\frac{kx_2+x_1}{k+1},\frac{ky_2+y_1}{k+1}\right),\qquad M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

Distance formulas (chapter recap)

$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\qquad OP=\sqrt{x^2+y^2}$$

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